Theoretical Mechanics - Babeș-Bolyai University

Lecture 8. Central forces 1

Theoretical Mechanics

10. Central forces.

Central forces. Properties (https://people.maths.bris.ac.uk/~marrk/Mech/15_16_CentralF_notes.pdf)

A central force is a force acting on a particle of mass m with the property that

the force is always directed from m toward, or away, from a fixed point O.

The particle is said to move in a central force field. The point O is referred to

as the centre of force. Examples:

Gravitational force

https://people.maths.bris.ac.uk/~marrk/Mech/15_16_CentralF_notes.pdf




















Examples:

Coulomb force

Elastic force(spring)

Attractive elastic force

Repulsive elastic force



Consider O the center of the force 𝐹 acting on the material point M(m) which

has the position vector 𝑟 = 𝑂𝑀.

Let be

the versor of 𝑟 . Thus, we have

(10.1)

where F is the algebraic

value of 𝐹 .

If F > 0 than 𝐹 is repulsive. Otherwise, if F < 0 than 𝐹 is attractive.



Using the moment of momentum theorem

(10.2)

one obtain the area first integral:

and, thus, the areal velocity is constant.

Remember: Areal velocity (sector velocity, sectorial velocity)

is the rate at which area is swept out by a particle as it

moves along a curve.

The areal velocity of M is:

(10.3)



From (10.2) and (10.3) we have:

In the motion under the action of the central force the moment of momentum

and the areal velocity are constant vectors at any moment 𝑡 ≥ 𝑡0.

(10.4)

We suppose:

Thus, the motion takes place in a plane determined by 𝑟0 and 𝑣 0 (actually the

normal to the plane is ).

From we have and then



Next, let us consider the motion of the point M(m) in the plane 𝑂𝑥𝑦 and let be

𝑟, 𝜃 the polar coordinates of the point M.

We have

(10.5)

But, and we obtain :

Equation (10.5) is the area integral (because is the area swept by M)

We have

and it means that

(10.6)



Remark. If 𝑐 = 0 then the motion of the point M is rectilinear.

Next we suppose

Taking into account the acceleration form in polar coordinate

(10.7)

Determination of motion in its plane

Consider the motion of the point M(m) in the plane 𝑂𝑥𝑦 and let be 𝑟, 𝜃 the

polar coordinates of the point M. The differential equation of motion is given by

and projecting (10.7) on the polar coordinates versors we obtain:



(10.8)

From (10.8)2 we obtain the area first integral

y

x

M0

M

er

e

F

v

O

r r0

v0 α

(10.9)

where

(10.10)



From (10.9) we have

And this means that the point M sweeps equal areas in equal intervals of time

(area law).

Thus, the motion of the material the point M

under the action of a central force respect

the area law and takes place in a plane (the

plane motion) determined by the initial

conditions.



Case I. 𝑭 = 𝑭(𝒓, 𝜽, 𝒓 , 𝜽 ) (i.e. 𝝏𝑭

𝝏𝒕= 𝟎)

One eliminate time 𝑡 from (10.8)1 by using the area integral (10.9):

Thus, Equation (10.8)1 becomes

(10.11)



When 𝐹 = 𝐹 𝑟, 𝜃 (i.e. The force depends only by position) the equation is as

the Binet‘s equation:

By solving eq. (10.12) one obtain the trajectory of the particle in polar

coordinates. However, in order to solve (10.12) two initial conditions are

necessary. From 𝑟 𝑡0 = 𝑟 0 and 𝑣 𝑡0 = 𝑣 0 we obtain:

(10.12)

(10.13)

Jacques Philippe Marie Binet

(February 2, 1786 – May 12, 1856)



Indeed,

Thus, we have

y

x

M0

M

er

e

F

v

O

r r0

v0 α



Solving (10.12) along with the initial conditions (10.13) one obtain the trajectory

Next, using the area integral r2 𝜃 = 𝑐 we have r2 𝜃 𝑑𝜃 = 𝑐𝑑𝑡 and we obtain:

(10.14)

(10.15)

Now, using (10.14) and (10.16) we get 𝑟 = 𝑟 𝑡 . In this moment the problem is

solved and the motion’s equations in polar coordinates are:

(10.16) and thus

(10.17)



Case II. 𝑭 = 𝑭(𝒓)

In this case it is possible to apply the general theorems of dynamics. First let us

calcuate the elementary work:

Thus, 𝛿𝐿 = 𝐹 𝑟 𝑑𝑟 is an exact differential

and the kinetic energy theorem dT = 𝛿𝐿 = −𝑑𝑉 becomes

𝛿𝐿 = −𝑑𝑉, 𝑉 ≔ − ∫ 𝐹 𝑟 𝑑𝑟

r

rdrrF

mvmvdrF

mvd

0

)(222

2

0

22

(10.18)



Thus,

hdrrFm

v )(22

Energy constant

2

2

22

2

2222

111

rrd

dcv

r

crd

dcr

rrv

We take into account that:

(10.19)

(10.20)



Thus,we have to solve

The energy constant can be calculated from

(10.21) hdrrF

mrrd

dc

)(

2112

2

2

hdrrFm

v )(22

0

(10.22)

Remark. It is also possible to use the Binet‘s equation for the case F=F(r).



Example 1.

A particle M(m) moves on a circle of radius b being attracted by a fixed point A

of the circle. Find the attractive force and the velocity of the particle as functions

of r, the distance between M and A.

Solution

Consider the equation of the circle

In polar coordinates we have



Next, in order to obtain the force, we calculate the left hand term of the equation

in order to obtain the force.

),(11

2

2

2

2

rFrrd

d

r

mc

Consider the Binet‘s equation

sign „+“ for repulsive force

sign „–“ for attractive force



Example 2.

Find the motion of a particle M(m=1) that moves under the action of an

attractive force 𝐹 𝑟 =1

𝑟3. At the initial moment we have:

𝑡 = 0: 𝜃0= 0, 𝑟0 = 2, 𝑣0 =1

2, 𝛼 = 𝑟 0, 𝑣 0

=𝜋

4

Solution



(equation of the trajectory in

polar coordinates, i.e. a

logarithmic spiral)



Using the area law

(equations of motion)



One can use matlab for visualization

t=0:0.01:100;

r=(sqrt(2)*t+4).^(1/2);

theta=(1/2)*log(sqrt(2)/4*t+1);

polar(theta,r)

Theoretical Mechanics - Babeș-Bolyai University

Documents

Transcript of Theoretical Mechanics - Babeș-Bolyai University